Which one of the following is the number by which the product of 8 consecutive integers is divisible?

## Concept & Rule A fundamental theorem in number theory states that the product of any $k$ consecutive integers is always perfectly divisible by $k!$ (k factorial). ## Step-by-Step Solution * The question asks about the product of exactly 8 consecutive integers. * Applying the fundamental rule, where $k = 8$, the product will always be divisible by $8!$. * We must now evaluate the options. The options list $4!$, $6!$, $7!$, and $8!$. * By definition, factorials are cumulative products: $$ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 $$ * Because $8!$ contains all the factors of the smaller factorials ($7! = 7 \times 6 \times \dots$, $6! = 6 \times 5 \times \dots$, etc.), any number divisible by $8!$ is automatically divisible by $7!$, $6!$, and $4!$. * Therefore, the product is divisible by every single factorial listed in the options. ## Exam Strategy & Shortcut Understand the hierarchy of divisibility. If a number is a multiple of a large factorial (like 8!), it is mathematically guaranteed to be a multiple of any smaller factorial. When you see a question format like this, and you identify the maximum factorial divisor, scan the options for an "All of these" choice. ## Common Pitfall A common mistake is reading the options sequentially, seeing that the product is divisible by 8!, and immediately selecting option (d) without reading option (e). Always read all available choices in multiple-choice questions, especially when factorials or multiples are involved. ## Final Answer **Therefore, the correct answer is All of these.**